How do you find the exact value of $arctan(tan ((-7pi)/6))$ ?

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Answer 1 · 2015-06-02T18:58:11

$tan(theta + n pi) = tan(theta)$ for any $n in ZZ$

So to make the definition of $arctan$ unique, its range is defined to be $(-pi/2, pi/2)$

$(-7pi)/6 + pi = -pi/6$ lies in the range of $arctan$ and is of the form $(-7pi)/6 + n pi$

So $arctan(tan((-7pi)/6)) = -pi/6$

How do you find the exact value of arctan(tan ((-7pi)/6)) arctan(tan ((-7pi)/6))?